I didn't find anywhere a simple function to describe a one-sheet hyperboloid based on arbitrary coordinates of the foci and the difference of distances between them.

So given the two points $F_1\, (x_1, y_1, z_1)$, $F_2\, (x_2, y_2, z_2)$ and the difference of distances $d \in \mathbb{R}$, I can define the hyperboloid as a locus of points $P(x,y,z)$ where $|PF_1| - |PF_2| = d$ that is

$\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2} - \sqrt{(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} = d$

The surface is closer to $F_1$ if $d<0$ and closer to $F_2$ if $d>0$. For $d=0$ it degenerates to a plane.

Is there a simpler form to describe this hyperboloid?

  • $\begingroup$ Technically, this looks to me like a hyperboloid of two sheets, even if you’re only interested in one of its components. It’s the surface of revolution of one branch of a hyperbola. $\endgroup$
    – amd
    Jan 31 '18 at 23:16
  • $\begingroup$ Thank you @amd, are you suggesting a parametrization? $\endgroup$
    – Guglie
    Jan 31 '18 at 23:19
  • $\begingroup$ I’d say construct a hyperbola in the $x$-$z$ plane with the foci on the $z$-axis, turn it into a surface of rotation by replacing $x^2$ with $x^2+y^2$ and then rotate and translate into position. I don’t see any pretty way to select only one branch of it without introducing square roots, though. That you could do more cleanly via parameterization. $\endgroup$
    – amd
    Jan 31 '18 at 23:23
  • $\begingroup$ Thanks @amd, I thought something similar, but I couldn't get it right, can you answer below so I can upvote you? $\endgroup$
    – Guglie
    Jan 31 '18 at 23:27

Your surface is one sheet of the surface generated by rotating a hyperbola about its transverse axis. The simplest representation is, I think, parametric. I would construct it in “standard position” first, and then rotate and translate it into place.

Let $f=|F_2-F_1|$. The hyperbola branch $x=\frac12\sqrt{f^2-d^2}\sinh t$, $z=\frac12 d\cosh t$ in the $x$-$z$ plane matches the given parameters, with the focus corresponding to $F_1$ on the negative $z$-axis. Rotating it about the $z$-axis turns this into $$x = \frac12\sqrt{f^2-d^2}\sinh t \cos\phi \\ y = \frac12\sqrt{f^2-d^2}\sinh t \sin\phi \\ z=\frac12 d\cosh t.$$ From here it’s a matter of rotating and translating it into place, which I’ll leave to you to work out. We really only care about the image of the $z$-axis, so taking advantage of the rotational symmetry it might be easier to construct a reflection that produces the desired tilt instead of rotating. This reflection would be in the angle bisector of the $z$-axis and $F_2-F_1$. (You might then want to replace $\phi$ with $-\phi$ in the parameterization to maintain orientation.) You can find some other parametric equations of a hyperbola to use as starting points here.

Alternatively, you could start with the implicit equation $${4z^2 \over d^2}-{4(x^2+y^2) \over f^2-d^2} = 1$$ of the above hyperbola and transform that, but I don’t see a good way to select one sheet of the surface without introducing square roots of the terms in the equation, which is what it appears you’re trying to avoid.


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